Theory of relativity - proton entering a medium

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Homework Help Overview

The discussion revolves around a problem related to the theory of relativity, specifically involving a proton entering a medium and the associated energy and momentum conservation principles. Participants are exploring the implications of given data such as the speed of light in different media and the mass of the proton.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using energy and momentum conservation equations to relate the initial and final states of the proton and the emitted photon. There are attempts to derive relationships between the momenta and energies involved, with some participants questioning the validity of their approximations and results.

Discussion Status

The discussion is ongoing, with various attempts to derive equations and clarify misunderstandings about the relationships between the variables. Some participants have provided guidance on how to approach the problem without approximations, while others express confusion about the implications of their calculations.

Contextual Notes

Participants note potential issues with dimensional consistency in their equations and the need to consider maximum and minimum values for the proton's momentum. There is also mention of constraints related to homework rules and the implications of setting certain variables to zero.

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Homework Statement


(see attachment)
Given data:
1.The speed of light in empty space is ##3 \times 10^8 m/s##.
2.The mass of the proton is ##1.67 \times 10^{-27} kg## .
3.The speed of light in water is ##2.3 \times 10^8 m/s##.
4.When photon emission begins to occur the initial proton, the final proton, and the emitted photon all travel in the same direction.

Homework Equations


The Attempt at a Solution


I don't have any knowledge about theory of relativity but I think this question can be done using the info given in the question.

Initial energy of proton, ##E_i=\sqrt{c^2p^2+m_0^2c^4}##
Final energy of proton, ##E_f=\sqrt{c^2p_1^2+m_0^2c^4}##
where ##m_0## is the rest mass of proton, ##p## is the initial momentum of proton and ##p_1## is the final momentum when it enters the water.
Using conservation of energy,
E_i=E_{photon}+E_f
Using conservation of linear momentum,
p=p_1+p_2
where ##p_2## is the momentum of emitted photon.
I can find ##p_2## in terms of ##p## and ##p_1## but I am not sure what I am supposed to do here. I can substitute p_2 in the energy equation but what am I supposed to do with that equation? :confused:

Any help is appreciated. Thanks!
 

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Anyone?
 
You've not used the equation Eph = cm|p2|.
Try approximating Ei, Ef on the basis that p << m0c.
 
Thanks Haruspex for your reply!
I used the equation E=cm|p|, took the approximation and substituted them in the energy equation. I ended up with a relation between p and p1.
p=2m_o^2c^2c_m-p_1
What am I supposed to do next?
 
Pranav-Arora said:
I used the equation E=cm|p|, took the approximation and substituted them in the energy equation. I ended up with a relation between p and p1.
p=2m_o^2c^2c_m-p_1
That can't be right - it's dimensionally inconsistent. (I get ##p=2m_oc_m-p_1##.)
How small can p1 be?
 
haruspex said:
That can't be right - it's dimensionally inconsistent. (I get ##p=2m_oc_m-p_1##.)
How small can p1 be?

Woops, I forgot to cancel ##m_oc^2## while taking the approximation. :redface:

smallest value of p1 can be zero?
 
Pranav-Arora said:
smallest value of p1 can be zero?
That's what I would assume.
 
ok so I plugged in the values. I get ##7.682 \times 10^{-19}## but this is wrong. :-(
 
Having ended up setting p1 to zero, I realize that we didn't need to do the approximation. Try putting p1=0 but not approximating anything. I think you should get 1.86 10-18. Is that better?
 
  • #10
haruspex said:
Having ended up setting p1 to zero, I realize that we didn't need to do the approximation. Try putting p1=0 but not approximating anything. I think you should get 1.86 10-18. Is that better?

It still says that its incorrect and now I can't try submitting an answer because all my tries are over. :-(
 
  • #11
Pranav-Arora said:
It still says that its incorrect and now I can't try submitting an answer because all my tries are over. :-(
And I'm out of ideas - sorry.
 
  • #12
haruspex said:
And I'm out of ideas - sorry.

The answers were published today. The answer given is ##3.58 \times 10^{-37}##.
 
  • #13
Pranav-Arora said:
The answers were published today. The answer given is ##3.58 \times 10^{-37}##.
Hmm.. that's numerically about the square of what we calculated. Any idea how that result is obtained?
EDIT: Just ran a check I wish I'd tried before: with my result, the implied initial speed of the proton > c. I guess setting p1=0 is not the way.
 
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  • #14
haruspex said:
I guess setting p1=0 is not the way.

Then how? :confused:
 
  • #15
Pranav-Arora said:
Then how? :confused:

My knowledge regarding general relativity is also meagre but the question asks the minimum magnitude of proton's momentum and not the maximum. You should rather choose for maximum value of p1. Its certainly not 0. What is it ?

Not quite sure of that. You can not be.

Edit: Aside, I really suspect that approximation will work.

Ok,ok. A more clear approach will be to eliminate one of the variables(not p) from momentum conservation , and the plug in energy conservation with two unknowns. Just isolate p in terms of p1 or p2 and analyze what value you can set for p1 or p2 for minimum p. What do you get ? Please do not approximate anything!

Really, you should get quadratic sort of equation.
 
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  • #16
sankalpmittal said:
My knowledge regarding general relativity is also meagre but the question asks the minimum magnitude of proton's momentum and not the maximum. You should rather choose for maximum value of p1. Its certainly not 0. What is it ?
p1 is the proton's momentum after the event. The approximation we tried to begin with suggested that the larger that was, the larger the initial momentum needed to be, so it seemed like the smallest p would go with p1=0. Having selected that, it was easy to rerun the calculation without approximating.
Ok,ok. A more clear approach will be to eliminate one of the variables(not p) from momentum conservation , and the plug in energy conservation with two unknowns. Just isolate p in terms of p1 or p2 and analyze what value you can set for p1 or p2 for minimum p. What do you get ? Please do not approximate anything!

Really, you should get quadratic sort of equation.
It gives a quartic, which is why I went for approximation. There might be some cancellation (of p-p1), but the algebra gets a bit heavy.
 
  • #17
haruspex said:
It gives a quartic, which is why I went for approximation. There might be some cancellation (of p-p1), but the algebra gets a bit heavy.

A quartic ? That will be bit messy. Ok, but you know that approximation was not giving the correct answer, right?

Can you state that quartic, in terms of p and p1 here? My exams are over on this very day and I am feeling too exhausted to do this. Or rather, we should wait for the OP to post... :rolleyes:
 
  • #18
sankalpmittal said:
A quartic ? That will be bit messy. Ok, but you know that approximation was not giving the correct answer, right?
It wasn't just wrong, it was out by about 18 orders of magnitude.
Can you state that quartic, in terms of p and p1 here? My exams are over on this very day and I am feeling too exhausted to do this. Or rather, we should wait for the OP to post... :rolleyes:
Initial energy of proton, ##E_i=\sqrt{c^2p^2+m_0^2c^4}##
Final energy of proton, ##E_f=\sqrt{c^2p_1^2+m_0^2c^4}##
where ##m_0## is the rest mass of proton, ##p## is the initial momentum of proton and ##p_1## is the final momentum when it enters the water.
Using conservation of energy,
E_i=E_{ph} +E_f
Using conservation of linear momentum, with p2 as the momentum of the photon
p=p_1+p_2
##E_{ph} = c_m|p_2|##
Putting it all together:
##\sqrt{c^2p^2+m_0^2c^4} - \sqrt{c^2p_1^2+m_0^2c^4} = c_m(p - p_1)##
Square both sides, isolate the surd on one side, square again. Actually it's not a quartic, but rather worse. OTOH, there may well be a factor of (p-p1) to take out.
 
  • #19
haruspex said:
It wasn't just wrong, it was out by about 18 orders of magnitude.

Initial energy of proton, ##E_i=\sqrt{c^2p^2+m_0^2c^4}##
Final energy of proton, ##E_f=\sqrt{c^2p_1^2+m_0^2c^4}##
where ##m_0## is the rest mass of proton, ##p## is the initial momentum of proton and ##p_1## is the final momentum when it enters the water.
Using conservation of energy,
E_i=E_{ph} +E_f
Using conservation of linear momentum, with p2 as the momentum of the photon
p=p_1+p_2
##E_{ph} = c_m|p_2|##
Putting it all together:
##\sqrt{c^2p^2+m_0^2c^4} - \sqrt{c^2p_1^2+m_0^2c^4} = c_m(p - p_1)##
Square both sides, isolate the surd on one side, square again. Actually it's not a quartic, but rather worse. OTOH, there may well be a factor of (p-p1) to take out.

Yes ! Thanks Haruspex ! :smile:

You are correct. I solved it further and the factor (p-p1) canceled out with the calculation. I have eliminated the radicals and have obtained the following equation :

c4(p+p1)2 + cm4(p-p1)2 - 2c2cm2p2= 2cm2c2p12 + 4cm2mo2c4

I guess that we will have mess with this equation to find minimum magnitude of p. Since all terms are basically squared, setting p1=0 , in this should do (but it did not ?).

There must be usage of some mathematical principle to find the magnitude of p1 so as to minimize p.

No, I think setting p1=0 should do. I get a quadratic equation in p.
I am sorry. There must be another way.

Nonetheless, no doubt this is the only way.
 
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  • #20
sankalpmittal said:
Yes ! Thanks Haruspex ! :smile:

You are correct. I solved it further and the factor (p-p1) canceled out with the calculation. I have eliminated the radicals and have obtained the following equation :

c4(p+p1)2 + cm4(p-p1)2 - 2c2cm2p2= 2cm2c2p12 + 4cm2mo2c4

I guess that we will have mess with this equation to find minimum magnitude of p. Since all terms are basically squared, setting p1=0 , in this should do (but it did not ?).

There must be usage of some mathematical principle to find the magnitude of p1 so as to minimize p.

No, I think setting p1=0 should do. I get a quadratic equation in p.

Nonetheless, no doubt this is the only way.

Setting p1=0 doesn't give the right answer. Oh my, how did you get the patience to do that dirty algebra. :-p
 
  • #21
Pranav-Arora said:
Setting p1=0 doesn't give the right answer. Oh my, how did you get the patience to do that dirty algebra. :-p

Then rather, open up the equation which I got in my previous post, and write it in the form : ap2+bp+c=0, where c contains all the terms of p1. The the discriminant b2-4ac≥0, should give you some sort of range of values of p1. And that was not very much algebra. Term p-p1 canceled out as marked by haruspex.
 
  • #22
sankalpmittal said:
Then rather, open up the equation which I got in my previous post, and write it in the form : ap2+bp+c=0, where c contains all the terms of p1. The the discriminant b2-4ac≥0, should give you some sort of range of values of p1. And that was not very much algebra. Term p-p1 canceled out as marked by haruspex.

I tried to differentiate the energy equation (consisting of surds) wrt ##p_1## to find the minimum value of p. I ended up with a quadratic in ##p_1##. Found two values of ##p_1# but none of them gave the right answer when I substituted them in the energy equation.
 
  • #23
Pranav-Arora said:
I tried to differentiate the energy equation (consisting of surds) wrt ##p_1## to find the minimum value of p. I ended up with a quadratic in ##p_1##. Found two values of ##p_1# but none of them gave the right answer when I substituted them in the energy equation.

Can you show the working? :-p

And please try the discriminant method which I cited in my previous post. I think it will give you the answer. Setting up a maxima or minima is sometimes problematic. Did you analyze d2p/dp12 also ?

Also if this does not work follow:

ap2+bp+c=0

Now use the quadratic formula and isolate p in terms of p1. One root of the two corresponding roots will be minimum value of p. Then from that root you can make p1 self adjust so that p will be surely minimum.

If all those method also not work (except that you do a careless mistake), then you will have to contact other HHs of PF. This is getting really messy now.
 
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  • #24
I'm not understanding something here. So, please help me out.

What if you go to the reference frame in which the proton is initially at rest. So, initially, the total energy in this frame is just the rest-mass energy of the proton. How can the proton in this frame emit a photon and conserve energy?

The only way that I can imagine the proton emitting a photon is if the proton interacts with the background water. But then you would need to include terms for the momentum and energy transferred to the water.
 
  • #25
TSny said:
I'm not understanding something here. So, please help me out.

What if you go to the reference frame in which the proton is initially at rest. So, initially, the total energy in this frame is just the rest-mass energy of the proton. How can the proton in this frame emit a photon and conserve energy?

The only way that I can imagine the proton emitting a photon is if the proton interacts with the background water. But then you would need to include terms for the momentum and energy transferred to the water.

Not only that, the very concept of "momentum" is controversial for photons moving in a medium. There are two conflicting definitions and over 100 years of debate which one is correct.
 
  • #26
TSny said:
I'm not understanding something here. So, please help me out.

What if you go to the reference frame in which the proton is initially at rest. So, initially, the total energy in this frame is just the rest-mass energy of the proton. How can the proton in this frame emit a photon and conserve energy?
In that frame, the proton sitting minding its own business got whacked by a tsunami. No shortage of energy there.
 
  • #27
Cracked it! It also occurred to me in the middle of the night that I should just differentiate the surd equation wrt p1 and set dp/dp1=0. That gives p = m0ccm/√(c2-cm2) = p1. At first sight that looks useless, since it gives p = p1, but then it dawned on me that the likely alternative to setting p1=0 is to set p-p1 very small. Indeed, that corresponds to the photon getting a vanishingly small amount of energy. (With that assumption in the first place, we could have expanded the surds on the basis of small δp = p-p1 and got the result quite quickly.)
Numerically it gives 5.98.10-19. Why so different from the given answer? Because the given answer is wrong, 3.58.10-37 being the numerical value of p2.
 
  • #28
haruspex said:
In that frame, the proton sitting minding its own business got whacked by a tsunami. No shortage of energy there.

OK, so the proton must interact with the water, and the "tsunami" transfers energy to the proton/photon system? How was that taken into account?
 
  • #29
For the record, I have tried a constrained minimization approach, and that got me zero momentum for the photon.
 
  • #30
Is the entire interaction of the proton with the water somehow taken into account by just using the dispersion relation E = cmp for the photon?
 

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